pn oi nhieu the nay ai ma giai cho het dc

bài lớp mấy mà nhìn ghê quá zật bạn..................Nhìu quá

sao mà trả ời hết đc

Ta có

xy + yz + xz \(\le\)x² + y² + z²

<=> 3(xy + yz + xz) \(\le\)(x + y + z)² = 9

<=> xy + yz + xz \(\le\)3

Vậy GTLN là 3 đạt được khi x = y = z = 1

sai rui

Vậy theo bạn đúng thì phải như thế nào :)

\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)

\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)

\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)

\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)

\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)

\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)

Dấu = xảy ra khi \(x=y=z=9\)

 Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\) 

CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)  ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\) 

Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)

\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) 

Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\) 

Mặt khác :   \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)

Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)

" = " \(\Leftrightarrow x=y=z=9\)

Nhân thêm và, dùng Cauchy

\(1\sqrt{x-1}=\sqrt{1\left(x-1\right)}\le\frac{x}{2}\). Tương tự với y thì nhân 2; với z thì nhân 3

\(\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)

\(=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

Ta có: \(\sqrt{x-1}\le\frac{1+x-1}{2}=\frac{x}{2}\)

\(\Rightarrow\frac{\sqrt{x-1}}{x}\le\frac{1}{2}\)

Chứng minh tương tự ta được: \(\frac{\sqrt{y-2}}{y}\le\frac{1}{2\sqrt{2}}\)

                                                 \(\frac{\sqrt{z-3}}{z}\le\frac{1}{2\sqrt{3}}\)

Suy ra: \(\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\le\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}\right)\)

Vậy GTLN của biểu thức = \(\frac{1}{2}.\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}\right)\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)

\(\frac{xy\sqrt{z-1}+xz\sqrt{y-2}+yz\sqrt{x-3}}{xyz}\\ =\frac{xy\sqrt{z-1}}{xyz}+\frac{xz\sqrt{y-2}}{xyz}+\frac{yz\sqrt{x-3}}{xyz}\\ =\frac{\sqrt{z-1}}{z}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{x-3}}{x}\\ =\frac{2\sqrt{z-1}}{2z}+\frac{2\sqrt{2}\sqrt{y-2}}{2\sqrt{2}y}+\frac{2\sqrt{3}\sqrt{x-3}}{2\sqrt{3}x}\)

Áp dụng BDT Cô-si với 2 số không âm:

\(\Rightarrow\frac{2\sqrt{z-1}}{2z}+\frac{2\sqrt{2}\sqrt{y-2}}{2\sqrt{2}y}+\frac{2\sqrt{3}\sqrt{x-3}}{2\sqrt{3}x}\\ \le\frac{1+\left(z-1\right)}{2z}+\frac{2+\left(y-2\right)}{2\sqrt{2}y}+\frac{3+\left(x-3\right)}{2\sqrt{3}x}\\ =\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}z-1=1\\y-2=2\\x-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=2\\y=4\\x=6\end{matrix}\right.\)

Vậy.......